I am trying to parse this page "https://www.imdb.com/title/tt0068112/?ref_=fn_al_tt_1", but I can't find the href that I need (href="/title/tt0068112/episodes?ref_=tt_eps_sm").
I tried with this code:
url="https://www.imdb.com/title/tt0068112/?ref_=fn_al_tt_1"
page(requests.get(url)
soup=BeautifulSoup(page.content,"html.parser")
for a in soup.find_all('a'):
print(a['href'])
What's wrong with this? I also tried to check "manually" with print(soup.prettify()) but it seems that that link is hidden or something like that.
You can get the page html with requests, the href item is in there, no need for special apis. I tried this and it worked:
import requests
from bs4 import BeautifulSoup
page = requests.get("https://www.imdb.com/title/tt0068112/?ref_=fn_al_tt_1")
soup = BeautifulSoup(page.content, "html.parser")
scooby_link = ""
for item in soup.findAll("a", href="/title/tt0068112/episodes?ref_=tt_eps_sm"):
print(item["href"])
scooby_link = "https://www.imdb.com" + "/title/tt0068112/episodes?ref_=tt_eps_sm"
print(scooby_link)
I'm assuming you also wanted to save the link to a variable for further scraping so I did that as well. 🙂
To get the link with Episodes you can use next example:
import requests
from bs4 import BeautifulSoup
url = "https://www.imdb.com/title/tt0068112/?ref_=fn_al_tt_1"
soup = BeautifulSoup(requests.get(url).content, "html.parser")
print(soup.select_one("a:-soup-contains(Episodes)")["href"])
Prints:
/title/tt0068112/episodes?ref_=tt_eps_sm
Related
from bs4 import BeautifulSoup
import requests
from urllib.request import urlopen
url = f'https://www.apple.com/kr/search/youtube?src=globalnav'
response = requests.get(url)
html = response.text
soup = BeautifulSoup(html, 'html.parser')
links = soup.select(".rf-serp-productname-list")
print(links)
I want to crawl through all links of shown apps. When I searched for a keyword, I thought links = soup.select(".rf-serp-productname-list") would work, but links list is empty.
What should I do?
Just check this code, I think is what you want:
import re
import requests
from bs4 import BeautifulSoup
pages = set()
def get_links(page_url):
global pages
pattern = re.compile("^(/)")
html = requests.get(f"your_URL{page_url}").text # fstrings require Python 3.6+
soup = BeautifulSoup(html, "html.parser")
for link in soup.find_all("a", href=pattern):
if "href" in link.attrs:
if link.attrs["href"] not in pages:
new_page = link.attrs["href"]
print(new_page)
pages.add(new_page)
get_links(new_page)
get_links("")
Source:
https://gist.github.com/AO8/f721b6736c8a4805e99e377e72d3edbf
You can change the part:
for link in soup.find_all("a", href=pattern):
#do something
To check for a keyword I think
You are cooking a soup so first at all taste it and check if everything you expect contains in it.
ResultSet of your selection is empty cause structure in response differs a bit from your expected one from the developer tools.
To get the list of links select more specific:
links = [a.get('href') for a in soup.select('a.icon')]
Output:
['https://apps.apple.com/kr/app/youtube/id544007664', 'https://apps.apple.com/kr/app/%EC%BF%A0%ED%8C%A1%ED%94%8C%EB%A0%88%EC%9D%B4/id1536885649', 'https://apps.apple.com/kr/app/youtube-music/id1017492454', 'https://apps.apple.com/kr/app/instagram/id389801252', 'https://apps.apple.com/kr/app/youtube-kids/id936971630', 'https://apps.apple.com/kr/app/youtube-studio/id888530356', 'https://apps.apple.com/kr/app/google-chrome/id535886823', 'https://apps.apple.com/kr/app/tiktok-%ED%8B%B1%ED%86%A1/id1235601864', 'https://apps.apple.com/kr/app/google/id284815942']
So the website I am using is : https://keithgalli.github.io/web-scraping/webpage.html and I want to extract all the social media links on the webpage.
import requests
from bs4 import BeautifulSoup as bs
r = requests.get('https://keithgalli.github.io/web-scraping/webpage.html')
soup = bs(r.content)
links = soup.find_all('a', {'class':'socials'})
actual_links = [link['href'] for link in links]
I get an error, specifically:
KeyError: 'href'
For a different example and webpage, I was able to use the same code to extract the webpage link but for some reason this time it is not working and I don't know why.
I also tried to see what the problem was specifically and it appears that
links is a nested array where links[0] outputs the entire content of the ul tag that has class=socials so its not iterable so to speak since the first element contains all the links rather than having each social li tag be seperate elements inside links
Here is the solution using css selectors:
import requests
from bs4 import BeautifulSoup as bs
r = requests.get('https://keithgalli.github.io/web-scraping/webpage.html')
soup = bs(r.content, 'lxml')
links = soup.select('ul.socials li a')
actual_links = [link['href'] for link in links]
print(actual_links)
Output:
['https://www.instagram.com/keithgalli/', 'https://twitter.com/keithgalli', 'https://www.linkedin.com/in/keithgalli/', 'https://www.tiktok.com/#keithgalli']
Why not try something like:
import requests
from bs4 import BeautifulSoup as bs
r = requests.get('https://keithgalli.github.io/web-
scraping/webpage.html')
soup = bs(r.content)
links = soup.find_all('a', {'class':'socials'})
actual_links = [link['href'] for link in links if 'href' in link.keys()]
After gaining some new information from you and visiting the webpage, I've realized that you did the following mistake:
The socials class is never used in any a-element and thus you won't find any such in your script. Instead you should look for the li-elements with the class "social".
Thus your code should look like:
import requests
from bs4 import BeautifulSoup as bs
r = requests.get('https://keithgalli.github.io/web-
scraping/webpage.html')
soup = bs(r.content, "lxml")
link_list_items = soup.find_all('li', {'class':'social'})
links = [item.find('a').get('href') for item in link_list_items]
print(links)
i just started programming.
I have the task to extract data from a HTML page to Excel.
Using Python 3.7.
My Problem is, that i have a website, whith more urls inside.
Behind these urls again more urls.
I need the data behind the third url.
My first Problem would be, how i can dictate the programm to choose only specific links from an ul rather then every ul on the page?
from bs4 import BeautifulSoup
import urllib
import requests
import re
page = urllib.request.urlopen("file").read()
soup = BeautifulSoup(page, "html.parser")
print(soup.prettify())
for link in soup.find_all("a", href=re.compile("katalog_")):
links= link.get("href")
if "katalog" in links:
for link in soup.find_all("a", href=re.compile("alle_")):
links = link.get("href")
print(soup.get_text())
There are many ways, one is to use "find_all" and try to be specific on the tags like "a" just like you did. If that's the only option, then use regular expression with your output. You can refer to this thread: Python BeautifulSoup Extract specific URLs. Also please show us either the link, or html structure of the links you want to extract. We would like to see the differences between the URLs.
PS: Sorry I can't make comments because of <50 reputation or I would have.
Updated answer based on understanding:
from bs4 import BeautifulSoup
import urllib
import requests
page = urllib.request.urlopen("https://www.bsi.bund.de/DE/Themen/ITGrundschutz/ITGrundschutzKompendium/itgrundschutzKompendium_node.html").read()
soup = BeautifulSoup(page, "html.parser")
for firstlink in soup.find_all("a",{"class":"RichTextIntLink NavNode"}):
firstlinks = firstlink.get("href")
if "bausteine" in firstlinks:
bausteinelinks = "https://www.bsi.bund.de/" + str(firstlinks.split(';')[0])
response = urllib.request.urlopen(bausteinelinks).read()
soup = BeautifulSoup(response, 'html.parser')
secondlink = "https://www.bsi.bund.de/" + str(((soup.find("a",{"class":"RichTextIntLink Basepage"})["href"]).split(';'))[0])
res = urllib.request.urlopen(secondlink).read()
soup = BeautifulSoup(res, 'html.parser')
listoftext = soup.find_all("div",{"id":"content"})
for text in listoftext:
print (text.text)
I'm using BeautifulSoup to parse code of this site and extract URL of the results. But when using find_all command I get an empty list as output. I checked manually the HTML code that I download from the site, and it contains the appropriate class.
If somebody could point out where I make a mistake or show a better solution I would be grateful!
from bs4 import BeautifulSoup
import requests
page = requests.get("https://www.awf.edu.pl/pracownik/wyszukiwarka-pracownikow?result_5251_result_page=3&queries_search_query=&category_kategorie=wydzia_wychowania_fizycznego&search_page_5251_submit_button=Szukaj¤t_result_page=1&results_per_page=20&submitted_search_category=&mode=results")
soup = BeautifulSoup(page.content, 'html.parser')
results = soup.find_all('div', class_ = 'search-item photo')
`
I've also tried to use this code below to just find all links on the site and then separate that what I need, but in this instance, I get only parent tag. if in tag 'a' is nested another tag 'a' it is skipped, and from documentation, I thought it also would be included in the output.
from bs4 import BeautifulSoup
import requests
page = requests.get("https://www.awf.edu.pl/pracownik/wyszukiwarka-pracownikow?result_5251_result_page=3&queries_search_query=&category_kategorie=wydzia_wychowania_fizycznego&search_page_5251_submit_button=Szukaj¤t_result_page=1&results_per_page=20&submitted_search_category=&mode=results")
soup = BeautifulSoup(page.content, 'html.parser')
results = soup.find_all('a')
BeautifulSoup can't find class that exists on webpage?
I found this answer to a similar question, but in my case, I can see the HTML code that I want to find in my console when I use print(soup.prettify())
the problem you are facing is linked to the way you are parsing page.content.
replace:
soup = BeautifulSoup(page.content, 'html.parser')
with:
soup = BeautifulSoup(page.content, 'lxml')
hope this helps.
I want to extract the link
/stocks/company_info/stock_news.php?sc_id=CHC&scat=&pageno=2&next=0&durationType=Y&Year=2018&duration=1&news_type=
from the html of the page
http://www.moneycontrol.com/company-article/piramalenterprises/news/PH05#PH05
The following is the code that is used
url_list = "http://www.moneycontrol.com/company-article/piramalenterprises/news/PH05#PH05"
html = requests.get(url_list)
soup = BeautifulSoup(html.text,'html.parser')
link = soup.find_all('a')
print(link)
using beautiful soup. How would I go about it, using find_all('a") doesn't return the required link in the returned html.
Please try this to get Exact Url you want.
import bs4 as bs
import requests
import re
sauce = requests.get('https://www.moneycontrol.com/stocks/company_info/stock_news.php?sc_id=CHC&durationType=Y&Year=2018')
soup = bs.BeautifulSoup(sauce.text, 'html.parser')
for a in soup.find_all('a', href=re.compile("company_info")):
# print(a['href'])
if 'pageno' in a['href']:
print(a['href'])
output:
/stocks/company_info/stock_news.php?sc_id=CHC&scat=&pageno=2&next=0&durationType=Y&Year=2018&duration=1&news_type=
/stocks/company_info/stock_news.php?sc_id=CHC&scat=&pageno=3&next=0&durationType=Y&Year=2018&duration=1&news_type=
You just have to use the get method to find the href attribute:
from bs4 import BeautifulSoup as soup
import requests
url_list = "http://www.moneycontrol.com/company-article/piramalenterprises/news/PH05#PH05"
html = requests.get(url_list)
page= soup(html.text,'html.parser')
link = page.find_all('a')
for l in link:
print(l.get('href'))